Cryptography Reference
In-Depth Information
y
i
)
2
g
1
(
x
) and
v
P
(
x
Hence,
x
−
x
i
=
(
y
−
−
x
i
)
=
2. Finally, the function (
x
−
x
i
) corre-
sponds to
x
−
x
i
z
z
=
x
z
−
x
i
on the projective curve
E
. Since
v
O
E
(
x/z
)
=−
2 it follows from part 2 of Lemma
7.4.14
that
v
O
E
(
x
−
x
i
)
=−
2. Hence, if
P
=
(
x
i
,y
i
) then, in all cases, div(
x
−
x
i
)
=
(
P
)
+
(
ι
(
P
))
−
2(
O
E
) and deg(div(
x
−
x
i
))
=
0.
Exercise
7.7.9
and Lemma
7.7.10
determine the divisor of certain functions, and in both
cases they turn out to have degree zero. This is not a coincidence. Indeed, we now state a
fundamental
5
result which motivates the definition of the divisor class group.
k
∈ k
(
C
)
∗
. Then
deg(div(
f
))
=
Theorem 7.7.11
Let C be a curve over
. Let f
0
.
Proof
See Theorem
8.3.14
.
(
C
)
∗
. The following are equivalent:
Corollary 7.7.12
Let C be a curve over
k
and let f
∈ k
1.
div(
f
)
≥
0
.
∈ k
∗
.
3.
div(
f
)
2. f
=
0
.
Proof
Certainly statement 2 implies statement 3 and 3 implies 1. So it suffices to prove
1 implies 2. Let
f
(
C
)
∗
be such that div(
f
)
∈ k
≥
0
.
Then
f
is regular everywhere, so
choose some
P
0
∈
C
(
k
) and define
h
=
f
−
f
(
P
0
)
∈ k
(
C
). Then
h
(
P
0
)
=
0. If
h
=
0 then
∈ k
∗
.To
f
is the constant function
f
(
P
0
) and, sinc
e
f
is defined over
k
, it follows that
f
complete the proof suppose that
h
0 by Theorem
7.7.11
it follows that
h
must have at least one pole. But then
f
has a pole, which contradicts
div(
f
)
=
0in
k
(
C
). Since deg(div(
h
))
=
≥
0.
(
C
)
∗
. Then
div(
f
)
Corollary 7.7.13
Let C be a curve over
k
. Let f,h
∈ k
=
div(
h
)
if and
=
∈ k
∗
.
only if f
ch for some c
Exercise 7.7.14
Prove Corollary
7.7.13
.
7.8 Divisor class group
(
C
)
∗
}
is a subgroup of Div
0
k
We have seen that Prin
k
(
C
)
(
C
). Hence, since
all the groups are Abelian, one can define the quotient group; we call this the divisor class
group. It is common to use the notation Pic for the divisor class group since the divisor
={
div(
f
):
f
∈ k
5
This innocent-looking fact is actually the hardest result in this chapter to prove. There are several accessible proofs of the general
result: Stichtenoth (Theorem I.4.11 of [
529
]; also see Moreno [
395
] Lemma 2.2) gives a proof based on “weak approximation” of
valuations and this is probably the simplest proof for a reader who has already got this far through the current topic; Fulton [
199
]
gives a proof for projective plane curves based on Bezout's theorem; Silverman [
505
], Shafarevich [
489
], Hartshorne [
252
]and
Lorenzini [
355
] all give proofs that boil down to ramification theory of
f
:
C
→ P
1
, and this is the argument we will give in
the next chapter.
